Riddle #678

logicmathprobability

The same birthday

What is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?
Only 23 people need to be in the room. Our first observation in solving this problem is the following: (the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0 What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations). With some simple re-arranging of the formula, we get: the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday) So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer. The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far. These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918 More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365) We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%. Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.
91.22 %
48 votes

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A fly in between trains

Two trains are traveling toward each other on the same track, each at 60 miles per hour. When they are exactly 120 miles apart, a fly takes off from the front of one of the trains, flying toward the other train at a constant rate of 100 miles per hour. When the fly reaches the other train, it instantly changes directions and starts flying toward the other train, still at 100 miles per hour. It keeps doing this back and forth until the trains finally collide. If you add up all the distances back and forth that the fly has travelled, how much total distance has the fly travelled when the trains finally collide?
The fly has travelled exactly 100 miles. We can figure this out using some simple math. Becuase the trains are 120 miles apart when the fly takes off, and are travelling at 60 mph each, they will collide in exactly 1 hour. This gives the fly exactly 1 hour of flying time, going at a speed of 100 miles per hour. Thus, the fly will travel 100 miles in this hour.
94.24 %
44 votes

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Red and blue marbles

Your friend shows you two jars, one with 100 red marbles in it, the other with 100 blue marbles in it. He proposes a game. He'll put the two jars behind his back and tell you to pick one of them at random. You'll then close your eyes, he'll hand you the jar you picked, and you'll pick a random marble from that jar. You win if the marble you pick is blue, and you lose otherwise. To give you the best shot at winning, your friend gives you the two jars before the game starts and says you can move the marbles around however you'd like, as long as all 200 marbles are in the 2 jars (that is, you can't throw any marbles away). How should you move the marbles around to give yourself the best chance of picking a blue marble?
Put one blue marble in one jar, and put the rest of the marbles in the other jar. This will give you just about a 75% chance of picking a blue marble.
94.11 %
43 votes

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Flipping Lockers

There are 1 million closed school lockers in a row, labeled 1 through 1,000,000. You first go through and flip every locker open. Then you go through and flip every other locker (locker 2, 4, 6, etc...). When you're done, all the even-numbered lockers are closed. You then go through and flip every third locker (3, 6, 9, etc...). "Flipping" mean you open it if it's closed, and close it if it's open. For example, as you go through this time, you close locker 3 (because it was still open after the previous run through), but you open locker 6, since you had closed it in the previous run through. Then you go through and flip every fourth locker (4, 8, 12, etc...), then every fifth locker (5, 10, 15, etc...), then every sixth locker (6, 12, 18, etc...) and so on. At the end, you're going through and flipping every 999,998th locker (which is just locker 999,998), then every 999,999th locker (which is just locker 999,999), and finally, every 1,000,000th locker (which is just locker 1,000,000). At the end of this, is locker 1,000,000 open or closed?
Locker 1,000,000 will be open. If you think about it, the number of times that each locker is flipped is equal to the number of factors it has. For example, locker 12 has factors 1, 2, 3, 4, 6, and 12, and will thus be flipped 6 times (it will end be flipped when you flip every one, every 2nd, every 3rd, every 4th, every 6th, and every 12th locker). It will end up closed, since flipping an even number of times will return it to its starting position. You can see that if a locker number has an even number of factors, it will end up closed. If it has an odd number of factors, it will end up open. As it turns out, the only types of numbers that have an odd number of factors are squares. This is because factors come in pairs, and for squares, one of those pairs is the square root, which is duplicated and thus doesn't count twice as a factor. For example, 12's factors are 1 x 12, 2 x 6, and 3 x 4 (6 total factors). On the other hand, 16's factors are 1 x 16, 2 x 8, and 4 x 4 (5 total factors). So lockers 1, 4, 9, 16, 25, etc... will all be open. Since 1,000,000 is a square number (1000 x 1000), it will be open as well.
94.11 %
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Mick and John were in a 100 meter race. When Mick crossed the finish line, John was only at the 90 meter mark. Mick suggested they run another race. This time, Mick would start ten meters behind the starting line. All other things being equal, will John win, lose, or will it be a tie in the second race?
John will lose again. In the second race, Mick started ten meters back. By the time John reaches the 90 meter mark, Mick will have caught up him. Therefore, the final ten meters will belong to the faster of the two. Since Mick is faster than John, he will win the final 10 meters and of course the race.
93.98 %
42 votes

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Ants on a Board

There are 100 ants on a board that is 1 meter long, each facing either left or right and walking at a pace of 1 meter per minute. The board is so narrow that the ants cannot pass each other; when two ants walk into each other, they each instantly turn around and continue walking in the opposite direction. When an ant reaches the end of the board, it falls off the edge. From the moment the ants start walking, what is the longest amount of time that could pass before all the ants have fallen off the plank? You can assume that each ant has infinitely small length.
The longest amount of time that could pass would be 1 minute. If you were looking at the board from the side and could only see the silhouettes of the board and the ants, then when two ants walked into each other and turned around, it would look to you as if the ants had walked right by each other. In fact, the effect of two ants walking into each other and then turning around is essentially the same as two ants walking past one another: we just have two ants at that point walking in opposite directions. So we can treat the board as if the ants are walking past each other. In this case, the longest any ant can be on the board is 1 minute (since the board is 1 meter long and the ants walk at 1 meter per minute). Thus, after 1 minute, all the ants will be off the board.
93.98 %
42 votes

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Moochers

A man who lives in Middletown has two girlfriends, one in Northtown and one in Southtown. Trains from the Middletown train station leave for Northtown once every hour. Separate trains from the station also leave for Southtown once every hour. No trains go to both Northtown and Southtown. Each day he gets to the Middletown train station at a completely random time and gets onto the first train that is going to either Northtown or Southtown, whichever comes first. After a few months, he realizes that he spends 80% of his days with his girlfriend from Northtown, and only 20% of his days with his girlfriend from Southtown. How could this be?
The train to Northtown leaves every hour, on the hour (9:00AM, 10:00AM, etc...). The train to Southtown leaves at 12 after the hour (9:12AM, 10:12AM, etc...). So there is only a 12/60 (1/5) chance that he will end up on the train to Southtown each day, since he will usually get to the station during the 48 minutes of each hour when the train to Northtown will be the next to come.
93.84 %
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Flying across the Atlantic

Two planes take off at the same exact moment. They are flying across the Atlantic. One leaves New York and is flying to Paris at 500 miles per hour. The other leaves Paris and is flying to New York at only 450 miles per hour. Which one will be closer to Paris when they meet?
They will both the same distance from Paris when they meet!
93.70 %
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Threedoors, one prize

You are on a gameshow and the host shows you three doors. Behind one door is a suitcase with $1 million in it, and behind the other two doors are sacks of coal. The host tells you to choose a door, and that the prize behind that door will be yours to keep. You point to one of the three doors. The host says, "Before we open the door you pointed to, I am going to open one of the other doors." He points to one of the other doors, and it swings open, revealing a sack of coal behind it. "Now I will give you a choice," the host tells you. "You can either stick with the door you originally chose, or you can choose to switch to the other unopened door." Should you switch doors, stick with your original choice, or does it not matter?
You should switch doors. There are 3 possibilities for the first door you picked: You picked the first wrong door - so if you switch, you win You picked the other wrong door - again, if you switch, you win You picked the correct door - if you switch, you lose Each of these cases are equally likely. So if you switch, there is a 2/3 chance that you will win (because there is a 2/3 chance that you are in one of the first two cases listed above), and a 1/3 chance you'll lose. So switching is a good idea. Another way to look at this is to imagine that you're on a similar game show, except with 100 doors. 99 of those doors have coal behind them, 1 has the money. The host tells you to pick a door, and you point to one, knowing almost certainly that you did not pick the correct one (there's only a 1 in 100 chance). Then the host opens 98 other doors, leave only the door you picked and one other door closed. We know that the host was forced to leave the door with money behind it closed, so it is almost definitely the door we did not pick initially, and we would be wise to switch.
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Using only and all the numbers 3, 3, 7, 7, along with the arithmetic operations +,-,*, and /, can you come up with a calculation that gives the number 24? No decimal points allowed. [For example, to get the number 14, we could do 3 * (7 - (7 / 3))]
7 * ((3 / 7) + 3) = 24
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Two words are anagrams if and only if they contain the exact same letters with the exact same frequency (for example, "name" and "mean" are anagrams, but "red" and "deer" are not). Given two strings S1 and S2, which each only contain the lowercase letters a through z, write a program to determine if S1 and S2 are anagrams. The program must have a running time of O(n + m), where n and m are the lengths of S1 and S2, respectively, and it must have O(1) (constant) space usage.
First create an array A of length 26, representing the counts of each letter of the alphabet, with each value initialized to 0. Iterate through each character in S1 and add 1 to the corresponding entry in A. Once this iteration is complete, A will contain the counts for the letters in S1. Then, iterate through each character in S2, and subtract 1 from each corresponding entry in A. Now, if the each entry in A is 0, then S1 and S2 are anagrams; otherwise, S1 and S2 aren't anagrams. Here is pseudocode for the procedure that was described: def areAnagrams(S1, S2) A = new Array(26) A.initializeValues(0) for each character in S1 arrayIndex = mapCharacterToNumber(character) //maps "a" to 0, "b" to 1, "c" to 2, etc... A[arrayIndex] += 1 end for each character in S2 arrayIndex = mapCharacterToNumber(character) A[arrayIndex] -= 1 end for (i = 0; i < 26; i++) if A[i] != 0 return false end end return true end
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